Suppose $C^1([0,1]^n)$ is the set of real-valued functions defined on $[0,1]^n$, whose derivative $\leq 1$ is continuous on $[0,1]^n$.
Define $$Y=\{ f \in C^1([0,1]^n) : f(0)=0 \}$$ Why $Y$ is a closed subspace of codimension $1$?
The question is taken from the proof of Lemma $15$.
$Y$ is the kernel of the surjective (and continuous) map
$$C^1([0,1]^n) \to \mathbb R, f \mapsto f(0).$$
In particular, $Y$ is closed (as a kernel) and of codimension $1$ (since $C^1([0,1]^n)/Y \cong \mathbb R$ is one-dimensional).